2. What is the difference between UNION and UNION ALL? Write the syntax for each.

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1 SQL Advanced Discussion Questions Solutions 1. The relational set operators UNION, INTERSECT, and MINUS work properly only if the relations are union-compatible. What does union-compatible mean, and how would you check for this condition? Union compatible means that the relations yield attributes with identical names and compatible data types. That is, the relation A(c1,c2,c3) and the relation B(c1,c2,c3) have union compatibility if the columns have the same names, are in the same order, and the columns have compatible data types. Compatible data types do not require that the attributes be exactly identical -- only that they are comparable. For example, VARCHAR(15) and CHAR(15) are comparable, as are NUMBER (3,0) and INTEGER, and so on. 2. What is the difference between UNION and UNION ALL? Write the syntax for each. UNION yields unique rows. In other words, UNION eliminates duplicates rows. On the other hand, a UNION ALL operator will yield all rows of both relations, including duplicates. Notice that for two rows to be duplicated, they must have the same values in all columns. To illustrate the difference between UNION and UNION ALL, let s assume two relations: A (ID, Name) with rows (1, Lake, 2, River, and 3, Ocean) and B (ID, Name) with rows (1, River, 2, Lake, and 3, Ocean). SELECT * FROM A UNION SELECT * FROM B will yield: ID Name 1 Lake 2 River 3 Ocean 1 River 2 Lake while SELECT * FROM A UNION ALL SELECT * FROM B will yield: 1

2 ID Name 1 Lake 2 River 3 Ocean 1 River 2 Lake 3 Ocean 3. Suppose that you have two tables, EMPLOYEE and EMPLOYEE_1. The EMPLOYEE table contains the records for three employees: Alice Cordoza, John Cretchakov, and Anne McDonald. The EMPLOYEE_1 table contains the records for employees John Cretchakov and Mary Chen. Given that information, what is the query output for the UNION query? (List the query output.) The query output will be: Alice Cordoza John Cretchakov Anne McDonald Mary Chen 4. Given the employee information in Question 3, what is the query output for the UNION ALL query? (List the query output.) The query output will be: Alice Cordoza John Cretchakov Anne McDonald John Cretchakov Mary Chen 5. Given the employee information in Question 3, what is the query output for the INTERSECT query? (List the query output.) The query output will be: John Cretchakov 6. Given the employee information in Question 3, what is the query output for the EXCEPT query? (List the query output.) This question can yield two different answers. If you use SELECT * FROM EMPLOYEE 2

3 EXCEPT SELECT * FROM EMPLOYEE_1 the answer is Alice Cordoza Ann McDonald If you use SELECT * FROM EMPLOYEE_1 EXCEPT SELECT * FROM EMPLOYEE the answer is Mary Chen 7. What is a CROSS JOIN? Give an example of its syntax. A CROSS JOIN is identical to the PRODUCT relational operator. The CROSS JOIN is also known as the Cartesian product of two tables. For example, if you have two tables, AGENT, with 10 rows and CUSTOMER, with 21 rows, the CROSS JOIN resulting set will have 210 rows and will include all of the columns from both tables. Syntax examples are: SELECT * FROM CUSTOMER CROSS JOIN AGENT; or SELECT * FROM CUSTOMER, AGENT If you do not specify a join condition when joining tables, the result will be a CROSS Join or PRODUCT operation. 8. Explain the three join types? LEFT [OUTER] JOIN RIGHT [OUTER] JOIN FULL [OUTER] JOIN The LEFT [OUTER] JOIN will yield all rows with matching values in the join columns, plus all of the unmatched rows from the left table. (The left table is the first table named in the FROM clause.) 3

4 The RIGHT [OUTER] JOIN will yield all rows with matching values in the join columns, plus all of the unmatched rows from the right table. (The right table is the second table named in the FROM clause.) The FULL [OUTER] JOIN will yield all rows with matching values in the join columns, plus all the unmatched rows from both tables named in the FROM clause. 9. What is a subquery, and what are its basic characteristics? A subquery is a query (expressed as a SELECT statement) that is located inside another query. The first SQL statement is known as the outer query, the second is known as the inner query or subquery. The inner query or subquery is normally executed first. The output of the inner query is used as the input for the outer query. A subquery is normally expressed inside parenthesis and can return zero, one, or more rows and each row can have only one column. A subquery can appear in many places in a SQL statement: as part of a FROM clause, to the right of a WHERE conditional expression, to the right of the IN clause, in a EXISTS operator, to the right of a HAVING clause conditional operator, in the attribute list of a SELECT clause. Examples of subqueries are: INSERT INTO PRODUCT SELECT * FROM P; DELETE FROM PRODUCT WHERE V_CODE IN (SELECT V_CODE FROM VENDOR WHERE V_AREACODE = 615 ); SELECT FROM WHERE V_CODE, V_NAME VENDOR V_CODE NOT IN (SELECT V_CODE FROM PRODUCT); 10. What is a correlated subquery? Give an example. A correlated subquery is subquery that executes once for each row in the outer query. This process is similar to the typical nested loop in a programming language. Contrast this type of subquery to the typical subquery that will execute the innermost subquery first, and then the next outer query until the last outer query is executed. That is, the typical subquery will execute in serial order, one after another, starting with the innermost subquery. In contrast, a correlated subquery will run the outer query first, and then it will run the inner subquery once for each row returned in the outer subquery. 4

5 For example, the following subquery will list all the product line sales in which the units sold value is greater than the average units sold value for that product (as opposed to the average for all products.) SELECT INV_NUMBER, P_CODE, LINE_UNITS FROM LINE LS WHERE LS.LINE_UNITS > (SELECT AVG(LINE_UNITS) FROM LINE LA WHERE LA.P_CODE = LS.P_CODE); The previous nested query will execute the inner subquery once to compute the average sold units for each product code returned by the outer query. 11. Suppose that a PRODUCT table contains two attributes, PROD_CODE and VEND_CODE. Those two attributes have values of ABC, 125, DEF, 124, GHI, 124, and JKL, 123, respectively. The VENDOR table contains a single attribute, VEND_CODE, with values 123, 124, 125, and 126, respectively. (The VEND_CODE attribute in the PRODUCT table is a foreign key to the VEND_CODE in the VENDOR table.) Given that information, what would be the query output (show values) for questions a d below? Because the common attribute is V_CODE, the output will only show the V_CODE values generated by the each query. a. A UNION query based on these two tables? 125,124,123,126 b. A UNION ALL query based on these two tables? 125,124,124,123,123,124,125,126 c. An INTERSECT query based on these two tables? 123,124,125 d. A MINUS query based on these two tables? If you use PRODUCT MINUS VENDOR, the output will be NULL If you use VENDOR MINUS PRODUCT, the output will be 126 5

6 Advanced Queries use pubs 1. Display a list of all titles and sales numbers in the titles and sales tables, including title that have no sales. Use a join. SELECT titles.title, sales.qty FROM titles LEFT OUTER JOIN sales ON titles.title_id = sales.title_id 2. Display a list of author s last name and all associated titles that the author has published sorted by the author s last name. Use a join. Save it as a View named: Published Authors. CREATE VIEW PublishedAuthors AS SELECT authors.au_lname, titles.title FROM authors, titleauthor,titles WHERE titleauthor.au_id = authors.au_id AND titles.title_id = titleauthor.title_id ORDER BY authors.au_lname 3. Using a subquery, display all the authors (show last and first name, zip) that receive a royalty of 100% and live in California. Save it as a view titled: AuthorsView. When creating the view rename the authors last name and first name as Last Name and First Name. Create View AuthorsView (Zip, last Name, First Name AS SELECT authors.zip, authors.au_lname, authors.au_fname FROM authors WHERE authors.au_id IN (SELECT titleuathor.au_id FROM titleauthor WHERE titleauthor.royaltyper = ) AND authors.state = CA 4. Display the stores that did not sell a title Is Anger the Enemy? Select stores.stor_name from stores where stor_id not in ( select stor_id from sales inner join titles on sales.title_id = titles.title_id where titles.title = 'Is Anger the Enemy?' ) 5. Display a list of store names for sales past (Order Date is greater than 1993) Display store name, and order date. 6

7 SELECT dbo.stores.stor_name, dbo.sales.ord_date FROM dbo.sales INNER JOIN dbo.stores ON dbo.sales.stor_id = dbo.stores.stor_id WHERE (YEAR(dbo.sales.ord_date) > 1993) 6. Display a list of titles for books sold in store name 'News & Brews'. Display store name, title, and order date. SELECT FROM WHERE dbo.titles.title, dbo.stores.stor_name, dbo.sales.ord_date dbo.sales INNER JOIN dbo.stores ON dbo.sales.stor_id = dbo.stores.stor_id INNER JOIN dbo.titles ON dbo.sales.title_id = dbo.titles.title_id (dbo.stores.stor_name = 'News & Brews') 7. List total sales (qty) by title. Display total quantity and title columns. SELECT SUM(dbo.sales.qty) AS Expr1, dbo.titles.title FROM dbo.sales INNER JOIN dbo.titles ON dbo.sales.title_id = dbo.titles.title_id GROUP BY dbo.titles.title 8. List total sales (qty) by type. Display total quantity and type columns. SELECT SUM(dbo.sales.qty) AS Expr1, dbo.titles.type FROM dbo.sales INNER JOIN dbo.titles ON dbo.sales.title_id = dbo.titles.title_id GROUP BY dbo.titles.type 9. List total sales (qty*price) by type. Display total $ value and type columns. SELECT SUM(dbo.sales.qty * dbo.titles.price) AS Expr1, dbo.titles.type FROM dbo.sales INNER JOIN dbo.titles ON dbo.sales.title_id = dbo.titles.title_id GROUP BY dbo.titles.type 10. Calculate the total number of types by publisher. Show publisher name and total count of types of books for each publisher. SELECT FROM COUNT(dbo.titles.type) AS Expr1, dbo.publishers.pub_name dbo.titles INNER JOIN dbo.publishers ON dbo.titles.pub_id = dbo.publishers.pub_id 7

8 GROUP BY dbo.publishers.pub_name 11. Show publisher names that do not have any type of book. Display publisher name only. SELECT FROM WHERE dbo.titles.type, dbo.publishers.pub_name dbo.titles RIGHT OUTER JOIN dbo.publishers ON dbo.titles.pub_id = dbo.publishers.pub_id (dbo.titles.type IS NULL) 8